Results 1 to 3 of 3

Math Help - Help with groups

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    Alabama
    Posts
    1

    Help with groups

    Hi
    My wife is trying to do the question below. I cannot help out and wondered if someone could take me through it.
    Thanks



    Attached Thumbnails Attached Thumbnails Help with groups-group.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,403
    Thanks
    1486
    Awards
    1

    Re: Help with groups

    Quote Originally Posted by Roger12 View Post
    Hi
    My wife is trying to do the question below. I cannot help out and wondered if someone could take me through it.
    Because a+b+1=b+a+1 we see a*b=b*a so it is commutative.
    (a*b)*c=(a*b)+c+1=(a+b+1)+c+1=a+(b+c+1)+1=a*(b*c) so it is associative.
    You wife needs to show that -1 is the operational identity.
    You wife needs to show that -a-2 is the operational inverse of a.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,158
    Thanks
    596

    Re: Help with groups

    verifying that -1 is the identity is rather easy one you "know" -1 is the identity. but how did Plato know it was?

    it involves a bit of "detective work". let's find out what the identity would have to be, if there actually is one.

    by definition, an identity for * is an element e of Z, with:

    a*e = e*a = a, for any element a of Z.

    since we know * is commutative (because of Plato's post above), it suffices to find e with a*e = a, for all a (in Z).

    so we write down an equation in Z, and "solve for e".

    if a*e = a, what this means in terms of the operations we're used to is:

    a+e+1 = a.

    we can, once we're "back in the ordinary integers", subtract a from both sides, to get:

    e+1 = 0, which makes it clear e = -1 (since that's the only integer e with e+1 = 0, or: if you like, subtract 1 from both sides).

    we can use "the same trick" to find out what a-1 has to be.

    by definition, a-1 is an element b of Z with a*b = b*a = e.

    now, we know what "e" is (see above), so we do pretty much what we did above: write a*b = e in terms of regular operations in Z, and "solve for b".

    a*b = e translates into:

    a+b+1 = -1

    subtract 1 from both sides:

    a+b = -2

    subtract a from both sides:

    b = -2-a (since this formula only depends on "a", we're good).

    my point being, it's often not profitable to just "guess" what the identity and/or inverse might be. often, with just a little bit of reasoning, we can figure out what they HAVE to be.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: March 30th 2012, 04:53 PM
  2. About minimal normal groups and subnormal groups
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: October 20th 2011, 01:53 PM
  3. Quotient Groups - Infinite Groups, finite orders
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: August 11th 2010, 07:07 AM
  4. free groups, finitely generated groups
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 23rd 2009, 03:31 AM
  5. Order of groups involving conjugates and abelian groups
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: February 5th 2009, 08:55 PM

Search Tags


/mathhelpforum @mathhelpforum